2020 AMC 8 Problems/Problem 5
Contents
- 1 Problem
- 2 Solution 1
- 3 Solution 2
- 4 Solution 3
- 5 Video Solution by NiuniuMaths (Easy to understand!)
- 6 Video Solution by Math-X (First understand the problem!!!)
- 7 Video Solution (🚀Very Fast🚀)
- 8 Video Solution by North America Math Contest go go go
- 9 Video Solution by WhyMath
- 10 Video Solution
- 11 Video Solution by Interstigation
- 12 Video Solution by STEMbreezy
- 13 See also
Problem
Three fourths of a pitcher is filled with pineapple juice. The pitcher is emptied by pouring an equal amount of juice into each of cups. What percent of the total capacity of the pitcher did each cup receive?
Solution 1
Each cup is filled with of the amount of juice in the pitcher, so the percentage is .
Solution 2
The pitcher is full, i.e. full. Therefore each cup receives percent of the total capacity.
Solution 3
Assume that the pitcher has a total capacity of ounces. Since it is filled three fourths with pineapple juice, it contains ounces of pineapple juice, which means that each cup will contain ounces of pineapple juice. Since the total capacity of the pitcher was ounces, it follows that each cup received of the total capacity of the pitcher, yielding as the answer.
Video Solution by NiuniuMaths (Easy to understand!)
https://www.youtube.com/watch?v=8hgK6rESdek&t=9s
~NiuniuMaths
Video Solution by Math-X (First understand the problem!!!)
https://youtu.be/UnVo6jZ3Wnk?si=dBVy_7cb8HQGTdBs&t=433
~Math-X
Video Solution (🚀Very Fast🚀)
~Education, the Study of Everything
Video Solution by North America Math Contest go go go
https://www.youtube.com/watch?v=bRzVP_caKrU ~North America Math Contest go go go
Video Solution by WhyMath
~savannahsolver
Video Solution
https://youtu.be/eSxzI8P9_h8 ~ The Learning Royal
Video Solution by Interstigation
https://youtu.be/YnwkBZTv5Fw?t=165
~Interstigation
Video Solution by STEMbreezy
https://youtu.be/L_vDc-i585o?list=PLFcinOE4FNL0TkI-_yKVEYyA_QCS9mBNS&t=205
~STEMbreezy
See also
2020 AMC 8 (Problems • Answer Key • Resources) | ||
Preceded by Problem 4 |
Followed by Problem 6 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | ||
All AJHSME/AMC 8 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.